WBJEE · Physics · Thermal Properties of Matter
\(A\) 10 W electric heater is used to heat a container filled with \(0.5 \mathrm{kg}\) of water. It is found that the temperature of water and the container rises by \(3^{\circ} \mathrm{K}\) in \(15 \mathrm{min}\). The container is then emptied, dried and filled with \(2 \mathrm{kg}\) of oil. The same heater now raises the temperature of container-oil system by \(2^{\circ} \mathrm{K}\) in \(20 \mathrm{min}\). Assuming that there is no heat loss in the process and the specific heat of water is \(4200 \mathrm{d} \mathrm{kg}^{-1} \mathrm{K}^{-1}\), the specific heat of oil in the same unit is equal to
- A \(1.50 \times 10^{3}\)
- B \(2.55 \times 10^{3}\)
- C \(3.00 \times 10^{3}\)
- D \(5.10 \times 10^{3}\)
Answer & Solution
Correct Answer
(B) \(2.55 \times 10^{3}\)
Step-by-step Solution
Detailed explanation
\(m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=\) Work done \(m_{1} s_{1} \Delta t+m_{2} s_{2} \Delta t=P_{1} t_{1}\) where \(m_{1}=0.5 \mathrm{kg}\) Specific heat \(s_{1}=4200 \mathrm{J} / \mathrm{kg}-\mathrm{K}\)…
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